The plane $2x-3y+6z-11=0$ makes an angle ${\sin }^{-1}$ ($a$) with $x-axis$. The value of a is equal to 

We have equation of plane as $2x-3y+6z-11=0$.

Normal to the plane is  $\overrightarrow{n}=2\widehat{i}-3\widehat{j}+6\widehat{k}$.

Also x-axis is along the vector $\overrightarrow{a}=\widehat{i}+0\widehat{j}+0\widehat{k}$.

According to the question, 

$\begin{array}{l}\sin \alpha =\frac{|\overrightarrow{a}\cdot \overrightarrow{n}|}{\left|\overrightarrow{a}\right|\overrightarrow{n}\mid }\\ =\frac{|\overrightarrow{i}\cdot (2\widehat{i}-3\widehat{j}+6\widehat{k}\left)\right|}{\sqrt{1}\sqrt{4+9+36}}\end{array}$

$=\frac{2}{7}$